Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

	public static void main(String[] args) { //Passes samples, but WA
		Scanner input = new Scanner(System.in);
		int N = input.nextInt();
		double[][] dp = new double[N+1][N+1];
		double[] p = new double[N+1];
		double answer = 0.0;
		double bug = 1.0; //keep track of tails' product
		for (int i = 1; i <= N; i++) {
			p[i] = input.nextDouble();
		}
		dp[1][1] = p[1];
		bug=1.0-p[1];
		for (int i = 2; i <= N; i++) { //Filling diagonals of DP table
			dp[i][i] = dp[i-1][i-1]*p[i];
		}
		for (int j = 2; j <= N; j++) { //Filling 1st row of DP table
			dp[1][j] = dp[1][j-1]*(1.0-p[j])+p[j]*(bug);
			bug*=(1.0-p[j]);
		}
		for (int i = 2; i < N; i++) { //dp[N][N] is already filled, last row not necessary
			for (int j = i+1; j <= N; j++) {
				dp[i][j]=dp[i][j-1]*(1.0-p[j])+dp[i-1][j-1]*p[j]; 
			}
		}
		/*for (int i = 1; i <= N; i++) {
			for (int j = 1; j <= N; j++) {
				System.out.print(dp[i][j]+" ");
			}
			System.out.print("\n");
		}*/
		for (int i = N/2+1; i <= N; i++) {
			answer+=dp[i][N];
		}
		System.out.println(answer);
	}
}